1. Physics
  2. Work, Energy and Power
  3. Work–Energy Theorem

Work, Energy and Power · JEE & NEET Physics

Work–Energy Theorem: notes and previous year questions

The net work of all forces equals the change in kinetic energy: when to use it, and the one-line patterns for stopping, bullets, springs, falls and rough slopes.

Work–Energy Theorem in short

  • The net work of all forces equals the change in kinetic energy.
  • Positive net work speeds a body up, negative slows it down, zero leaves its speed unchanged.
  • Add the work of every force, keeping each sign.
  • Use the theorem for speeds and distances; use Newton's laws for times, accelerations and forces at an instant.

1The theorem

The work–energy theorem says that the net work done on a body equals the change in its kinetic energy. It links forces directly to speeds, without finding the acceleration or the time.

Wnet=ΔKE=12mvf2−12mvi2W_{\text{net}} = \Delta KE = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2W_net is the work of every force added together. It holds for steady and changing forces.
Net workSpeedExample
positivegoes uppushing a cart
negativegoes downbraking a car
zerostays the samea stone whirled at a steady speed

Example: a net work of +36 J on a 2 kg body at rest gives 36=12(2)v236 = \tfrac{1}{2}(2)v^2, so v=6v = 6 m/s.

2Add every force

The key word is net: include the work of every force, with its sign.

  1. Pick the start and the end. "Starts from rest" means vi=0v_i = 0; "comes to rest" means vf=0v_f = 0.
  2. List every force and the work it does, with its sign.
  3. Add them: W1+W2+W3+⋯=12mvf2−12mvi2W_1 + W_2 + W_3 + \dots = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2.
  4. Solve for what you want.
ForceIts workSign
gravitymghmgh+ going down, − going up
kinetic friction−μNs-\mu N susually −
applied forceFscos⁡θFs\cos\thetaeither
normal force (fixed surface)0zero
spring12kxi2−12kxf2\tfrac{1}{2}kx_i^2 - \tfrac{1}{2}kx_f^2either

3When to use it

The theorem knows nothing about time. Choose it when a question asks for a speed or a distance. Use Newton's laws and kinematics for a time, an acceleration, or a force at one instant.

The question asks forUse
a final or initial speedwork–energy
a distance to stopwork–energy
an accelerationNewton's second law
the time takenNewton's laws and kinematics
a force at one instantNewton's second law

4Pattern 1: friction stops a block

A block of mass mm slides at vv and friction stops it in a distance ss. Friction's work removes all the kinetic energy:

−μmg s=0−12mv2  ⇒  μ=v22gs-\mu mg\,s = 0 - \tfrac{1}{2}mv^2 \;\Rightarrow\; \mu = \frac{v^2}{2gs}

The mass cancels, and the stopping distance grows as v2v^2: twice the speed, four times the distance.

5Patterns 2 to 4: bullets, springs and falls

A bullet stopped in a depth dd. The average resisting force does work −Fd-Fd, which removes all of 12mv2\tfrac{1}{2}mv^2:

F=mv22dF = \frac{mv^2}{2d}

A 10 g bullet at 400 m/s stopped in 10 cm: 12mv2=800\tfrac{1}{2}mv^2 = 800 J, so F=800/0.1=8000F = 800/0.1 = 8000 N. A 20 g bullet at 300 m/s stopped in 15 cm: 900 J, so F=6000F = 6000 N.

A spring launch. A spring of constant kk squeezed by xx pushes a block of mass mm on a smooth floor. All of 12kx2\tfrac{1}{2}kx^2 becomes 12mv2\tfrac{1}{2}mv^2:

v=xkmv = x\sqrt{\frac{k}{m}}

With k=800k = 800 N/m and m=2m = 2 kg, a launch at 4 m/s needs x=vm/k=4×0.05=0.2x = v\sqrt{m/k} = 4 \times 0.05 = 0.2 m.

A fall from height hh (or a slide down any smooth ramp): mgh=12mv2mgh = \tfrac{1}{2}mv^2, so v=2ghv = \sqrt{2gh}. From 5 m, v=10v = 10 m/s, whatever the shape of the smooth ramp.

6Down a rough slope

A block slides from rest down a slope of height hh and angle θ\theta with friction μ\mu. The slope is h/sin⁡θh/\sin\theta long, and friction is μmgcos⁡θ\mu mg\cos\theta.

mgh−μmgcos⁡θ⋅hsin⁡θ=12mv2mgh - \mu mg\cos\theta \cdot \frac{h}{\sin\theta} = \tfrac{1}{2}mv^2
v=2gh (1−μcot⁡θ)v = \sqrt{2gh\,(1 - \mu\cot\theta)}

If μ=tan⁡θ\mu = \tan\theta, then 1−μcot⁡θ=01 - \mu\cot\theta = 0: friction's work cancels gravity's at every step, so a block released from rest gains no speed, and a pushed block slides at a steady speed.

7A changing force

The theorem works for changing forces too: find the work from the area under the FF–xx graph, then the speed from the work.

8Three traps

Example: a 2 kg block lowered 1 m at constant speed. Its speed does not change, so the net work is zero: gravity does +20 J and the tension −20 J.

Summary

Key ideas

  • The net work of all forces equals the change in kinetic energy.
  • Positive net work speeds a body up, negative slows it down, zero leaves its speed unchanged.
  • Add the work of every force, keeping each sign.
  • Use the theorem for speeds and distances; use Newton's laws for times, accelerations and forces at an instant.
  • Friction stopping a block gives μ = v²/2gs; stopping distance grows as v².
  • A bullet stopped in a depth d feels an average force mv²/2d.
  • A spring launch gives v = x√(k/m); a fall or smooth slide from h gives v = √(2gh).
  • On a rough slope, v = √(2gh(1 − μ cot θ)).
  • For a changing force, find the work from the area, then the speed.
  • The normal force can do work on moving surfaces; tension's work cancels only for the pair; spring work is 12kxi2−12kxf2\tfrac{1}{2}kx_i^2 - \tfrac{1}{2}kx_f^2.

Every equation

Work–energy theorem
Wnet=12mvf2−12mvi2W_{\text{net}} = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2
Net work
Wnet=W1+W2+W3+…W_{\text{net}} = W_1 + W_2 + W_3 + \dots
Friction stops a block
μ=v22gs\mu = \frac{v^2}{2gs}
Bullet stopped in depth d
F=mv22dF = \frac{mv^2}{2d}
Spring launch
v=xk/mv = x\sqrt{k/m}
Fall from height h
v=2ghv = \sqrt{2gh}
Rough slope, from rest
v=2gh(1−μcot⁡θ)v = \sqrt{2gh(1 - \mu\cot\theta)}
Spring's work
Ws=12kxi2−12kxf2W_s = \tfrac{1}{2}kx_i^2 - \tfrac{1}{2}kx_f^2
Changing force
W=∫F dxW = \int F\,dx

Previous year questions with solutions

Real JEE and NEET questions on work–energy theorem. Try each one before you open the solution.

Q1JEE Main 2026One correct option

A spherical ball of mass 2 kg falls from a height of 10 m and is brought to rest after penetrating 10 cm into sand.

The average force exerted by sand on the ball is _________ N.

(Take g=10m/s2g=10 m/s^{2})

  1. A1980
  2. B2020
  3. C2000
  4. D1000
Show answer and solution

Answer: Option B

Take the whole trip at once, from release to rest in the sand: the ball starts and ends at rest, so ΔK=0\Delta K = 0 and the works of the two forces must cancel. Gravity acts the whole way down, through h+d=10.1 mh + d = 10.1\ \mathrm{m}, doing mg(h+d)=2×10×10.1=202 Jmg(h + d) = 2 \times 10 \times 10.1 = 202\ \mathrm{J}. The sand acts only over the last d=0.1 md = 0.1\ \mathrm{m}, doing −Favg×0.1-F_{\mathrm{avg}} \times 0.1. So Favg=2020.1=2020 NF_{\mathrm{avg}} = \dfrac{202}{0.1} = 2020\ \mathrm{N}. 2000 N2000\ \mathrm{N} is the answer if you forget that gravity keeps pulling during the 10 cm10\ \mathrm{cm} in the sand; 1980 N1980\ \mathrm{N} subtracts that bit instead of adding it.

Q2NEET 2026One correct option

A particle of mass MM moves along a horizontal xx axis from x=0x=0 to x=Lx=L. The coefficient of kinetic friction varies as a function of xx as μk(x)=μ0−αx{\mu}_{k}(x)={\mu}_{0}-\alpha x, where μ0{\mu}_{0}, α\alpha are constants of appropriate dimensions, so that μk(L)=0{\mu}_{k}(L)=0. The magnitude of the total work done by the frictional force during the motion is nμ0MgLn{\mu}_{0}MgL, where gg is the acceleration due to gravity. The value of nn is:

  1. A12\frac{1}{2}
  2. B3
  3. C1
  4. D13\frac{1}{3}
Show answer and solution

Answer: Option A

Since μk(L)=0\mu_{k}(L) = 0, αL=μ0\alpha L = \mu_{0}. The friction is μkMg=(μ0−αx)Mg\mu_{k}Mg = (\mu_{0} - \alpha x)Mg, and its size summed over the path is ∫0L(μ0−αx)Mg dx=Mg(μ0L−αL22)=Mg(μ0L−μ0L2)=12μ0MgL\displaystyle\int_{0}^{L}(\mu_{0} - \alpha x)Mg\,dx = Mg\left(\mu_{0}L - \dfrac{\alpha L^{2}}{2}\right) = Mg\left(\mu_{0}L - \dfrac{\mu_{0}L}{2}\right) = \tfrac{1}{2}\mu_{0}MgL. So n=12n = \tfrac{1}{2}. On a graph it is simply a triangle: the friction falls in a straight line from μ0Mg\mu_{0}Mg to zero over the length LL. The work done BY friction is −12μ0MgL-\tfrac{1}{2}\mu_{0}MgL, since it opposes the motion; the question asks only for its size. Option C keeps μ\mu at μ0\mu_{0} all the way, as though the floor never got smoother.

Q3JEE Advanced 2010One correct option

A block of mass 2 kg is free to move along the x-axis. It is at rest and from t = 0 onwards, it is subjected to a time-dependent force F(t) in the x-direction. The force F(t) varies with t along a single straight line: it is 4 N at t = 0, falls to 0 at t = 3 s, and carries on falling at the same rate to −2 N at t = 4.5 s. The kinetic energy of the block after 4.5 s is

  1. A4.50 J
  2. B7.50 J
  3. C5.06 J
  4. D14.06 J
Show answer and solution

Answer: Option C

The impulse is the area under the force–time graph, counted positive above the axis and negative below it. From 00 to 3 s3\ \mathrm{s} the area is a triangle, 12×3×4=6 N s\tfrac{1}{2} \times 3 \times 4 = 6\ \mathrm{N\,s}. From 33 to 4.5 s4.5\ \mathrm{s} the force is negative, a triangle of 12×1.5×2=1.5 N s\tfrac{1}{2} \times 1.5 \times 2 = 1.5\ \mathrm{N\,s} below the axis. Starting from rest, p=6−1.5=4.5 kg m/sp = 6 - 1.5 = 4.5\ \mathrm{kg\,m/s}, so K=p22m=4.524=20.254≈5.06 JK = \dfrac{p^{2}}{2m} = \dfrac{4.5^{2}}{4} = \dfrac{20.25}{4} \approx 5.06\ \mathrm{J}. Forget that the last stretch pulls backwards and you get p=7.5p = 7.5 and 14.06 J14.06\ \mathrm{J}; stop at t=3 st = 3\ \mathrm{s} and you get 9 J9\ \mathrm{J}, which is not even offered.

Practice questions, easy to hard

Three questions from the work–energy theorem practice ladder: one easy, one medium, one hard.

Q4One correct option

Because K=12mv2K = \tfrac{1}{2}mv^{2} squares the speed, kinetic energy grows much faster than speed does: twice the speed means 22=42^{2} = 4 times the kinetic energy. The same squaring works with momentum. For one body the mass is fixed, so K=p22mK = \dfrac{p^{2}}{2m} makes K∝p2K \propto p^{2}. Percentage changes then compound rather than add.

The momentum of a body goes up by 20%20\%. By what percentage does its kinetic energy go up?

  1. A44%44\%
  2. B20%20\%
  3. C40%40\%
  4. D400%400\%
Show answer and solution

Answer: Option A

pp becomes 1.2p1.2p, so KK becomes (1.2)2K=1.44K(1.2)^{2}K = 1.44K — a rise of 44%44\%. 40%40\% is the quick answer of doubling the 20%20\%, which works only for very small changes; it misses the extra 4%4\% that comes from 0.2×0.20.2 \times 0.2. Always turn the percentage into a factor, square or square-root the factor, and turn it back.

Q5One correct option

A sum of terms is undone one term at a time, and a constant term aa is undone to axax.

A force F=4+6xF = 4 + 6x (in N\mathrm{N}, with xx in m\mathrm{m}) moves a body along +x+x from x=0x = 0 to x=2 mx = 2\ \mathrm{m}. How much work does it do?

  1. A32 J32\ \mathrm{J}
  2. B12 J12\ \mathrm{J}
  3. C16 J16\ \mathrm{J}
  4. D20 J20\ \mathrm{J}
Show answer and solution

Answer: Option D

G=4x+3x2G = 4x + 3x^{2}, so W=G(2)−G(0)=8+12=20 JW = G(2) - G(0) = 8 + 12 = 20\ \mathrm{J}. The graph is a straight line from 4 N4\ \mathrm{N} to 16 N16\ \mathrm{N}, and the trapezium agrees: an average of 10 N10\ \mathrm{N} over 2 m2\ \mathrm{m}. A is the final force times the distance; B drops the constant 44, as if a steady part of a force did no work; C is F(2)F(2) itself — a force, not a work.

Q6Numerical answer

Now bring back ∫F dx\int F\,dx. A force that depends on position does work you can compute by undoing a derivative, and the theorem turns that work into a change of speed.

A 2 kg2\ \mathrm{kg} block moves along the xx-axis at 4 m/s4\ \mathrm{m/s}. As it passes x=0x = 0 it enters a region where the only horizontal force on it is F=−6x2 NF = -6x^{2}\ \mathrm{N} (xx in m\mathrm{m}). How far past x=0x = 0, in m\mathrm{m}, does it travel before it stops?

Show answer and solution

Answer: 2 m

Over a distance dd the force does W=∫0d−6x2 dx=−2d3W = \int_{0}^{d} -6x^{2}\,dx = -2d^{3}, negative because it points against the motion. The block arrives with 12×2×42=16 J\tfrac{1}{2} \times 2 \times 4^{2} = 16\ \mathrm{J} and stops when all of it is gone: 2d3=162d^{3} = 16, so d3=8d^{3} = 8 and d=2 md = 2\ \mathrm{m}. There is no constant acceleration here, so v2=u2+2asv^{2} = u^{2} + 2as cannot be used — the theorem works regardless, because work adds up correctly whatever the force does along the way.